Sizing a solar system for a village fish freezer
A coastal village wants to run a 150 W chest freezer for fish. The compressor runs for about 8 hours in every 24 hours. The village has one 300 W solar panel, a charge controller and a 12 V, 100 Ah lead-acid battery. Average peak sun is 5 hours per day, and the supplier says to allow 30% losses in the panel, wiring and controller. The inverter is 90% efficient. The freezer uses about 0.6 kWh of AC electricity between sunset and sunrise. The battery should not be discharged below 50% to keep it healthy. A second panel costs K500 and a second battery costs K850. The village has K4,000, but also needs a K800 inverter and a K400 controller. Cloudy weather can last two days, and the freezer must not thaw.
1. Calculate the freezer's daily energy demand and the usable energy from one panel. Decide how many panels are needed. Show your working.
Read the worked answer
Daily demand: 150 W × 8 h = 1,200 Wh = 1.2 kWh per day.
One panel: 300 W × 5 peak sun hours = 1,500 Wh = 1.5 kWh DC per day. With 30% losses, usable energy = 1.5 × 0.70 = 1.05 kWh per day.
One panel supplies only 1.05 kWh but the freezer needs 1.2 kWh, so one panel is not enough. Number needed = 1.2 ÷ 1.05 = 1.14, and panels come in whole units, so the village needs 2 panels. Two panels give about 2.1 kWh per day, which covers the load and leaves some margin for cloudy days. Rounding down to one panel would leave the freezer short every day.
2. Calculate the usable AC energy from one battery and decide whether one battery can supply the 0.6 kWh night load. Explain what happens over two cloudy days.
Read the worked answer
One battery: 12 V × 100 Ah = 1,200 Wh = 1.2 kWh stored. At 50% depth of discharge, usable DC energy = 600 Wh. After the inverter (90% efficient), usable AC energy = 600 × 0.90 = 540 Wh = 0.54 kWh.
The night load is 0.6 kWh AC, which is greater than 0.54 kWh. So one battery is not enough; it would be discharged below 50% and its life would shorten. Two batteries give 1,080 Wh AC usable, enough for one night with a small margin.
Over two cloudy days, the panels may produce very little, so the batteries would not recharge. Two batteries cover only about one night, not two days. The village would need to reduce load, add more battery capacity, or have a backup such as a small generator.
3. Using the costs in the setup, decide the best purchase within K4,000. Show the total and justify why you would not simply buy the cheapest option.
Read the worked answer
Best combination: 2 panels + 2 batteries + inverter + controller.
- Panels: 2 × K500 = K1,000
- Batteries: 2 × K850 = K1,700
- Inverter: K800
- Controller: K400
- Total = K3,900, leaving K100.
The cheapest option, 1 panel + 1 battery + inverter + controller, costs K2,550, but it fails both tests: one panel supplies only 1.05 kWh/day against a 1.2 kWh demand, and one battery supplies only 0.54 kWh AC against a 0.6 kWh night load. Three panels and two batteries would cost K4,400, over budget. The 2-panel, 2-battery system meets the daily and night requirements and leaves a small reserve. The village should still manage load carefully during cloudy spells and plan to add battery capacity later.